Answer
$\displaystyle \color{blue}{\frac{3e^{-3t} + 2e^{5t}}{5}}$
Work Step by Step
$\underline{\text{The Sample Space for the Draws}}$
Note that a chip cannot be drawn twice as the draws are without replacement.
${\begin{array}{cc}
\underline{(C_1,C_2)} & \rm Chip\ 2\ (C_2) \\
\rm Chip\ 1\ (C_1)& \begin{array}{|c|c|c|c|c|} \hline
C_1\backslash C_2& 1 & 2 & 3 & 4 & 5 \\ \hline
1 & & (1,2) & (1,3) & (1,4) & (1,5) \\
2 & (2,1) & & (2,3) & (2,4) & (2,5) \\
3 & (3,1) & (3,2) & & (3,4) & (3,5) \\
4 & (4,1) & (4,2) & (4,3) & & (4,5) \\
5 & (5,1) & (5,2) & (5,3) & (5,4) & \\ \hline
\end{array}
\end{array}}$
The possible values of the sum of the numbers on the two chips that were randomly drawn.
${\begin{array}{cc}
\underline{C_1+C_2} & \rm Chip\ 2\ (C_2) \\
\rm Chip\ 1\ (C_1)& \begin{array}{|c|c|c|c|c|} \hline
+ & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & & 3 & 4 & 5 & 6 \\
2 & 3 & & 5 & 6 & 7 \\
3 & 4 & 5 & & 7 & 8 \\
4 & 5 & 6 & 7 & & 9 \\
5 & 6 & 7 & 8 & 9 & \\ \hline
\end{array}
\end{array}}$
$\underline{{\rm The\ Random\ Variable}\ X}$
Since $X$ is $5$ if the sum is odd and $-3$ if the sum is even, the value of $X$ for each possible sum of the numbers on the two chips that were randomly drawn must be:
${\begin{array}{cc}
\underline{C_1+C_2} & \rm Chip\ 2\ (C_2) \\
\rm Chip\ 1\ (C_1)& \begin{array}{|c|c|c|c|c|} \hline
+ & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & & -3 & 5 & -3 & 5 \\
2 & -3 & & -3 & 5 & -3 \\
3 & 5 & -3 & & -3 & 5 \\
4 & -3 & 5 & -3 & & -3 \\
5 & 5 & -3 & 5 & -3 & \\ \hline
\end{array}
\end{array}}$
Note that $12$ of the $20$ sums are $-3$ and $8$ of the $20$ sums are $5$.
Thus, based on all the possible outcomes for the draw, the proportion of $-3$'s is $12/20 = 3/5$ and the proportion of $5$'s is $8/20=2/5$.
The pdf of $X$ is, therefore,
$f_X(x) = \begin{cases} \dfrac{3}{5}, & x=-3 \\ \dfrac{2}{5}, & x=5 \\ 0, & \text{elsewhere}. \end{cases}$
$\underline{{\rm The\ mgf\ of}\ X}$
$\begin{align*}
M_X(t) &= E(e^{Xt}) \\
&= \sum_{-\infty}^\infty e^{xt}f_X(x)\ \\
&= \sum\limits_{x\in\{-3,5\}} e^{xt}f_X(x) \qquad \text{[ since}\ f_X(x)=0, x\not\in \{-3,5\}\ ] \\
&= e^{-3t}f_X(-3) + e^{5t}f_X(5) \\
&= e^{-3t}\cdot \frac{3}{5} + e^{5t}\cdot\frac{2}{5} \\
\color{blue}{M_X(t)}\ &\color{blue}{=\frac{3e^{-3t} + 2e^{5t}}{5}}
\end{align*}$
